Discreet Mathamatics

[Flikerator]

I started Discreet Math a few days ago, and I have an assignment due (The class is 2/3rds done, I started late, its in E-Course) monday. Its the last question I have a problem with.

T or F, the vectors ( 8, 2 , 4 ), ( 3 , -9 , 6) , ( -5 , 10 , 0) form a basis in R3.

I found that they do not (I later tried it a few times in different ways, same result; not coplanar).

The bonus question says, write (#,#,#) as a linear combination of the vectors in the previous question. I replaced the numbers with # because I don't want anyone doing the work for me.

If the three vectors do not form a basis in R3, how can I write them as a combination of another vector? Unless it wants me to sub each the three equations in for the new one.

Any ideas?

[Flikerator]

Also is (A [dot] A) U the same as (A [dot]A) [dot] U?

Dot = Dot Product

[bugzpodder]

a,b,c are LI iff (a x b) . c = 0

just solve u*a + v*b + w*c = (#, #, #), three equations and three unknowns


dpeends on what u mean by A and U

[Brightguy]

bugzpodder wrote:

a,b,c are LI iff (a x b) . c = 0

Doh! The determinant of a matrix is 0 when the matrix is singular, i.e. linearly dependent columns/rows.

Flikerator wrote:

I found that they do not (I later tried it a few times in different ways, same result; not coplanar).

A basis for R3 is formed by any set of 3 linearly independent vectors in R3. So, if the vectors are not coplanar then.......

Flikerator wrote:

Also is (A [dot] A) U the same as (A [dot]A) [dot] U?

Besides the trivial case when U is a scalar, these aren't the same. The second isn't even defined unless U is a scalar.

[Flikerator]

Thank you both very much, it helped a lot.

Also, I'm doing projection and I came into a snag. "5. Find the shortest distance from the point P(#, #, #) to the line which passes through C(#, #,#) and D(#,#, #)."

This is my procedure and reasoning. I think I made a mistake here somewhere;

Find the direction vector of CD (C-D)
Find the direction vector of PD (could be PC)
Use PD and CD to find a perpendicular line to the vectors. (Cross product)
Use the cross product to form a Cartesian equation (need a normal vector, thats why I got the CP).
Solve for D (Ax + By + Cz + D = 0)

Then plug the numbers in for projection;

The problem I think is that I used P to form a vector to find the perpindicular vector. Even though I used point C to find D in the CE, it still works out to the same thing if I use P.

So when I have;
P(x, y, z)

|Ax + By + Cz + D|
-----------------------
sqrt(A² + B² + C²)

I get 0 on the top. So the perpendicular line must pass through point P (if its true).

Is that just coincidence that it worked out that way, or what did I do wrong? Im pretty sure its wrong, and it might be early on when I'm declaring which vectors (by using P).



Note: I havn't done Proj in R3. We learned R2 and now have to do one in R3 =/

[Brightguy]

The equation of the plane isn't necessary, you just need to know projection. Informally, the projection of x onto y is like dropping a perpendicular from x onto y. The projection is then the new vector along y defined by where the dropped perpendicular hits y.

So, you start by finding CD and CP (or similar). Notice that the perpendicular dropped in proj(CP onto CD) is the most direct vector from P to the CD line - that's the one you want to find.

Let proj(CP onto CD) = CE, so the vector we want to find is EP. But since CE + EP = CP, we have EP = CP - CE. Then just find the length of EP.

(It would help to draw a diagram...)

[bugzpodder]

axb.c is the triple cross product which calculates the volume of parallelpiped in 3space. a,b,c are coplanar iff volume is 0.

Flikerator wrote:

|Ax + By + Cz + D|
-----------------------
sqrt(A² + B² + C²)

thats the distance from (x,y,z) to the plane Au+Bv+Cw+D = 0

for ur original problem, just do:

|PC x CD|/|CD| since PC x CD = |PC||CD|cos(t) where t is the angle between them. Draw this on a 2-d plane if u cant figure out the 3d space, since any 3 points can be normalized onto a single plane.

[Flikerator]

I drew it out and it makes a lot of sense now. Thank you for the help its really appreciated.

So if EP = CP - CE, then CE also equals CD because they lie on the same line?

My teacher isn't going to be online until tuesday, could someone verify if this answers correct? I want to make sure I get it.

code:

P(11, 2, 5) C(-3, 1,-7) D(5,3, 5) EP = CP - CE CP = C-P = (-3, 1, 7) - (11, 2, 5) = (-3 - 11, 1 - 2, 7 - 5) = (-14, -1, 2) = (14, 1, -2) CE = CD = C-D = (-3, 1, 7) - (5, 3, 5) = (-3 - 5, 1 - 3, 7 - 5) = (-8, -2, 2) = (4, 1, -1) EP = CP - CE       = (14, 1, -2) - (4, 1, -1)       = (14 - 4, 1 - 1, -2 - (-1))       = (10, 0, -1)

So the distance from the line CD to the point P, is the vector (10, 0, -1)?


------------------------
This has me highly confused.
The only way for vectors to add to the zero vector when they are coplanar is when the scalars are all equal to 0. Thus they are LI.
Vectors are LD when they are coplanar and add to the zero vector. So if they arent coplanar, they are LI. If they dont add to the 0 vector (all scalars are 0) they are LI.

To find if three (or more) vectors are coplanar (LD), they would be a linear combination of the rest of the vectors. So if the values of t and k (scalars) are the same for all the equations, then it is LD and coplanar.

However, if this is not true, they are LI and NOT coplanar. If they are LI and not coplanar, then they form a basis? But the isn't the definition of a coplanar; they form a plane?

So if they ARENT coplanar, they form plane. But if the vectors are coplanar, they dont form a plane?...

EDIT: I think I get it. The VECTORS are not coplanar; but the points they make can be coplanar (With only three vectors they are). So if they points they make can form planes; then the non coplanar vectors can form a basis. I'm pretty sure thats it.

----------
And for division of a line segement. Is the ratio -1:2 the same as 2:-1? I assume not, but assumptions have got me wrong answers before.

[bugzpodder]

this is highly confusing. search for linear independence on google to get better definitions.

[Brightguy]

Flikerator wrote:

So if EP = CP - CE, then CE also equals CD because they lie on the same line?

No, CE and CD will be collinear, but will usually not be the same. It's the same in 2 dimensions, proj(x onto y) will usually not equal y.

Flikerator wrote:

code:

(-8, -2, 2) = (4, 1, -1)

Woah, these vectors are not equal! Since one has been multiplied by a scalar they are collinear, but not equal.

Also, CD = D - C, not C - D. To distinguish between points and vectors, perhaps this is written more clearly as CD = OD - OC, with O the origin, i.e. the point (0, 0, 0).

There are multiple ways to approach the question, of course. You can also just plug the points into the point-line distance formula that bugzpodder's was referring to. It's derived (with a bit of calculus) on this page. In the diagram on that page, the point E that I introduced would lie at the intersection of the solid and dotted line.

Flikerator wrote:

The only way for vectors to add to the zero vector when they are coplanar is when the scalars are all equal to 0. Thus they are LI.

If the vectors are coplanar, all the scalars do not have to equal 0 to sum to the zero vector, thus they are linearly dependent.

Here's a summary of linear independence for R3:

• One vector is always linearly independent, unless it's the zero vector.
  
• Two vectors are always linearly independent, unless they are collinear.
  
• Three vectors are always linearly independent, unless they are coplanar.
  
• Four or more vectors are never linearly independent.
  

As you pointed out, you have to distinguish between coplanar points and vectors. Three points in R3 are always coplanar.

Flikerator wrote:

And for division of a line segement. Is the ratio -1:2 the same as 2:-1? I assume not, but assumptions have got me wrong answers before.

Right, not the same.

bugzpodder wrote:

axb.c is the triple cross product which calculates the volume of parallelpiped in 3space. a,b,c are coplanar iff volume is 0.

Coplanar is right (you said LI originally). Also, you mean |PC×CD| = |PC||CD|sin(t).

[Flikerator]

Shankz you very much. Im onto equations of lines/planes now, but its still really appreciated